Question

What is the largest number that divides 236 and 596, leaving remainder 11 in each case?​

Answer 1

Answer:

Step-by-step explanation:

we have to  find the largest number which divide 615 and 963 leaving remainder 6 in each case.

so let us subtract 6 from 615 and 963

⇒ 615−6=609

⇒ 963−6=957

now lets find the HCF

⇒ prime factorisation of 609=29×3×3

prime factorisation of 957=29×3×11

now lets take out the common factors from both the cases

⇒ 29 and 3

x=29×3=87

∴ 87 is the number which will divide 615 and 963 leaving remainder 6 in each case.